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# exam1solution_pdf - Version 444 – Exam 1 – Sutcliffe...

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Unformatted text preview: Version 444 – Exam 1 – Sutcliffe – (52440) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points For gases that do not react chemically with water, the solubility of the gas in water gen- erally (decreases, increases) with an increase in the pressure of the gas and (decreases, in- creases) with increasing temperature. 1. decreases; decreases 2. decreases; increases 3. increases; decreases correct 4. increases; increases Explanation: An increase in pressure means that you have increased the concentration of gas above the solvent surface, thereby increasing the concentration of the gas in the solvent. In- creasing the temperature will decrease the solubility of the gas. 002 10.0 points Calculate the molarity of a solution prepared by dissolving 11 . 5 g NaOH in enough water to make 1 . 5 L of solution. 1. 0.43 M 2. 2.3 M 3. 7.6 M 4. 0.19 M correct Explanation: m NaOH = 11 . 5 g V = 1 . 5 L Molarity is moles solute per liter of solution. We use the molar mass of NaOH (40 g/mol) to convert from grams to moles NaOH: mol NaOH = 11 . 5 g NaOH × 1 mol NaOH 40 g NaOH = 0 . 2875 mol NaOH We divide the moles of NaOH in the solution by the total volume of the solution: M = . 2875 mol NaOH 1 . 5 L solution = 0 . 19 M NaOH 003 10.0 points Note: The question is referring to finding the equilibrium constant for the forward reaction. The standard molar Gibbs free energy of formation of NO 2 (g) at 298 K is 51.30 kJ · mol − 1 and that of N 2 O 4 (g) is 97.82 kJ · mol − 1 . What is the equilibrium constant at 25 ◦ C for the reaction 2 NO 2 (g) ⇀ ↽ N 2 O 4 (g) ? 1. 6.88 correct 2. 9 . 72 × 10 9 3. None of these 4. 7 . 01 × 10 − 9 5. 1.00 6. 0.657 7. 0.145 8. 1 . 02 × 10 − 10 Explanation: Δ G products = 97 . 82 kJ · mol − 1 Δ G reactants = 51 . 30 kJ · mol − 1 Δ G rxn = summationdisplay n Δ G products- summationdisplay n Δ G reactants = 97 . 82- (2)(51 . 30) = (- 4 . 78 kJ / mol) parenleftbigg 1000 J kJ parenrightbigg =- 4780 J / mol Δ G =- RT ln K K = e − Δ G / ( R T ) = exp bracketleftbigg-- 4780 J / mol (8 . 3145 J / mol · K)(298 K) bracketrightbigg = 6 . 88395 Version 444 – Exam 1 – Sutcliffe – (52440) 2 004 10.0 points The vapor pressures of pure carbon disulfide and carbon tetrachloride are 360 and 99.8 torr, respectively, at 296 K. What is the vapor pressure of a solution containing 50.0 g of each compound? 1. 460 torr 2. 33.0 torr 3. 241 torr 4. 260 torr 5. 274 torr correct Explanation: 005 10.0 points A certain reaction has Δ H equal to 11.57 kJ/mol. This reaction is normally run at room temperature (25 ◦ C). At what new tem- perature should the reaction be run so that K is twice its value at 25 ◦ C?...
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## This note was uploaded on 10/25/2010 for the course CH 302 taught by Professor Holcombe during the Fall '07 term at University of Texas.

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exam1solution_pdf - Version 444 – Exam 1 – Sutcliffe...

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